2.
Each of the graphs below contains two curves.
Identify the graph containing a curve representing a function y = f ( x ) and a curve
representing its second derivative y = f ( x ) .
(A)
(B)
(C)
(D)
(E)
May 2003
7
Course 1
1.
A survey of a groups viewing habits over the last year revealed the following
information:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
28% watched gymnastics
29% watched baseball
19% watched soccer
14% watched gymnastics and baseball
12% watched baseball and so
10.
E
Let
W = event that wife survives at least 10 years
H = event that husband survives at least 10 years
B = benefit paid
P = profit from selling policies
Then
and
Pr [ H ] = P [ H W ] + Pr H W c = 0.96 + 0.01 = 0.97
Pr [W | H ] =
Pr [W H ]
Pr W | H =
c
21.
E
The differential equation that we are given is separable. As a result, the general solution
is given by
1
Q ( N Q ) dQ = dt = t + C
where C is a constant. Now in order to calculate the integral on the lefthand side of this
equation, we first need t
19.
B
Let X1, Xn denote the life spans of the n light bulbs purchased. Since these random
variables are independent and normally distributed with mean 3 and variance 1, the
random variable S = X1 + + Xn is also normally distributed with mean
= 3n
and sta
13.
E
Let
X = number of group 1 participants that complete the study.
Y = number of group 2 participants that complete the study.
Now we are given that X and Y are independent.
Therefore,
P ( X 9 ) (Y < 9 ) ( X < 9 ) (Y 9 )
cfw_
= P ( X 9 ) (Y < 9 ) + P
11.
D
Observe that x and y follow the constraint equation
x + y = 160, 000
x = 160, 000 y where 0 y 160, 000
Now this constraint equation can be used to express policy sales g(x, y) as a function f(y)
of marketing y alone:
f ( y ) = g (160, 000 y, y ) = 0
38.
D
From f , observe that
4 x + c1 for 0 < x < 10
f ( x ) = kx + c2 for 10 < x < 30
3 x + c for x > 30
3
As a result, 200 = f(50) = 3(50) + c3 = 150 + c3 implies c3 = 50
And 0 = f ( 0 ) = 4 ( 0 ) + c1 = c1 ,
Then due to the continuity requirement,
10k +
26.
B
Let
u be annual claims,
v be annual premiums,
g(u, v) be the joint density function of U and V,
f(x) be the density function of X, and
F(x) be the distribution function of X.
Then since U and V are independent,
1
1
g (u , v ) = ( e u ) e v / 2 = e
30.
B
Let
x = number of ice cream cones sold
p(x) = price of x ice cream cones
C(x) = cost of selling x ice cream cones
R(x) = revenue from selling x ice cream cones
P(x) = profit from selling x ice cream cones
We are told that p(x) satisfies the followin
17.
B
Let Y denote the claim payment made by the insurance company.
Then
with probability 0.94
0
Y = Max ( 0, x 1) with probability 0.04
14
with probability 0.02
and
E [Y ] = ( 0.94 )(0 ) + ( 0.04 )( 0.5003)
15
1
( x 1) e x / 2 dx + (0.02 )(14 )
15
15
=
28.
C
Since f (t ) > 0
and f ' (t ) < 0
(i ) f (t0 ) > f (t1 )
for t 0 , the following inequalities hold:
if
0 t0 < t1
(ii ) f ( k ) < k 1 f (t )dt
if k 1
k
k +1
(iii ) f ( k ) > k
f (t )dt
k 0
if
Applying these inequalities, we see that
f ( 0 ) + f (1) +
33.
D
Let
IA = Event that Company A makes a claim
IB = Event that Company B makes a claim
XA = Expense paid to Company A if claims are made
XB = Expense paid to Company B if claims are made
Then we want to find
Pr I AC I B ( I A I B ) ( X A < X B )
cfw_
27.
A
First, observe that the distribution function of X is given by
x 3
1
1
F ( x ) = 4 dt = 3 |1x = 1 3 , x > 1
1 t
t
x
Next, let X1, X2, and X3 denote the three claims made that have this distribution. Then if
Y denotes the largest of these three claim
22.
C
Let X denote the waiting time for a first claim from a good driver, and let Y denote the
waiting time for a first claim from a bad driver. The problem statement implies that the
respective distribution functions for X and Y are
F ( x ) = 1 e x / 6 ,
28.
Note a more heuristic approach to the result that (E) > (B) > (A) > (C) >
20
f ( k ) > (D)
k =1
can be obtained from diagrams of the following sort:
20
20
2
1
gives ( E ) = f (0 ) + f (1) + f (t ) dt > f ( 0 ) + f (t ) dt = ( B )
and
20
20
1
k =1
giv
11.
Alternate solution using Lagrange multipliers:
Solve:
x + y 160, 000 = 0
1
3
1
3
x 4y 4
= ( x + y 160, 000 )
x 1000
x
x 4y 4
= ( x + y 160, 000 )
y 1000
y
From last two equations:
3
3
1
x 4y 4 =
4000
1
1
3
x 4y 4 =
4000
Eliminating :
1
1
3
3x 4 y 4
24.
E
The domain of s and t is pictured below.
Note that the shaded region is the portion of the domain of s and t over which the device
fails sometime during the first half hour. Therefore,
1/ 2 1
1 1/ 2
1
1
Pr S T = f ( s, t ) dsdt + f ( s, t ) dsdt
0
15.
D
At the point (0, 5),
0 = 2t 2 + t 1 = ( 2t 1)(t + 1)
5 = t 2 3t + 1
1
The first equation says t = or t = 1 and the second says t = 1 .
2
The slope of the tangent line to C at (0, 5) is then
dy
2t 3
dy
dx
=
=
dx ( x , y )=(0,5) dt t =1 dt t =1 4t + 1
31.
C
A Venn diagram for this situation looks like:
We want to find w = 1 ( x + y + z )
1
1
5
We have x + y = , x + z = , y + z =
4
3
12
Adding these three equations gives
1 1 5
(x + y) + (x + z) + ( y + z) = + +
4 3 12
2(x + y + z) = 1
x+ y+ z =
1
2
1 1
5.
C
The domain of X and Y is pictured below. The shaded region is the portion of the domain
over which X<0.2 .
Now observe
1 x
0.2
1 2
6
1
x
+
y
dydx
=
6
(
)
0 y xy 2 y 0 dx
0 0
0.2
0.2
1
1
2
2
2
= 6 1 x x (1 x ) (1 x ) dx = 6 (1 x ) (1 x ) dx
0
0
2
2
1.
E
We are given that
5e 5b = p5 = p6 = 6e 6b
It follows that
5 e 6 b
= 5b = e 6 b e5 b = e 6 b+5 b = e b
6 e
5
ln = b
6
5
6
b = ln = ln
6
5
2.
C
First, solve for m such that
500 = 8 + 8 (1.07 ) + . + 8 (1.07 )
m 1
1 (1.07 )m
(1.07 )m 1
= 8
= 8
1 1
Course 1
May 2001
Answer Key
1
2
3
4
5
E
C
C
E
C
21
22
23
24
25
E
C
A
E
B
6
7
8
9
10
D
B
A
C
E
26
27
28
29
30
B
A
C
D
B
11
12
13
14
15
D
D
E
A
D
31
32
33
34
35
C
E
D
A
C
16
17
18
19
20
A
B
D
B
D
36
37
38
39
40
B
E
D
A
B
A
B
C
D
E
Course 1 Solutions
7
8
8
9.
C
The Venn diagram below summarizes the unconditional probabilities described in the
problem.
In addition, we are told that
P[A B C]
1
x
= P [ A B C | A B] =
=
3
P [ A B]
x + 0.12
It follows that
1
1
x = ( x + 0.12 ) = x + 0.04
3
3
2
x = 0.04
3
x = 0.0
4.
E
Let X1, X2, X3, and X4 denote the four independent bids with common distribution
function F. Then if we define Y = max (X1, X2, X3, X4), the distribution function G of Y is
given by
G ( y ) = Pr [Y y ]
= Pr ( X 1 y ) ( X 2 y ) ( X 3 y ) ( X 4 y )
= P
COURSE 1
MAY 2001
2.
A stock pays annual dividends. The first dividend is 8 and each dividend thereafter is 7%
larger than the prior dividend.
Let m be the number of dividends paid by the stock when the cumulative amount paid
first exceeds 500 .
Calculate
1.
The price of an investment at the end of month n is modeled by pn = nebn where b is a
constant. The model predicts that the price at the end of the sixth month is the same as
the price at the end of the fifth month.
Determine b .
May 2001
(A)
5
ln
6
(
1.
The price of an investment at the end of month n is modeled by pn = nebn where b is a
constant. The model predicts that the price at the end of the sixth month is the same as
the price at the end of the fifth month.
Determine b .
May 2001
(A)
5
ln
6
(